Understanding the Floating Cube Problem

This problem involves a solid cube floating in water. We are asked to find the volume of the cube that remains outside the water surface. To solve this, we need to apply the principles of buoyancy and density, specifically Archimedes' principle.

Given Information:

• Mass of the solid cube, \(m = 800 \text{ gm}\)
• Edge length of the cube, \(L = 10 \text{ cm}\)
• Density of water, \(\rho_{\text{water}} = 1000 \text{ kg/m}^3\)

Step-by-Step Solution:

1. Calculate the total volume of the cube:

The volume of a cube is given by the edge length cubed.

Volume of cube, \(V_{\text{total}} = L^3\)

\(V_{\text{total}} = (10 \text{ cm})^3\)

\(V_{\text{total}} = 1000 \text{ cm}^3\)

2. Convert units for consistency (optional but good practice):

Let's convert the mass of the cube and the density of water to consistent units, like grams and cm³ or kilograms and m³. Using grams and cm³ is often easier for this type of problem when density is given in gm/cm³ or easily convertible.

• Mass of cube, \(m = 800 \text{ gm}\)
• Density of water, \(\rho_{\text{water}} = 1000 \text{ kg/m}^3\). We know that \(1 \text{ kg} = 1000 \text{ gm}\) and \(1 \text{ m} = 100 \text{ cm}\), so \(1 \text{ m}^3 = (100 \text{ cm})^3 = 10^6 \text{ cm}^3\).
• \(\rho_{\text{water}} = \frac{1000 \text{ kg}}{1 \text{ m}^3} = \frac{1000 \times 1000 \text{ gm}}{10^6 \text{ cm}^3} = \frac{10^6 \text{ gm}}{10^6 \text{ cm}^3} = 1 \text{ gm/cm}^3\)

So, Density of water, \(\rho_{\text{water}} = 1 \text{ gm/cm}^3\)

3. Apply Archimedes' Principle for Floating Objects:

When an object floats, the buoyant force acting upwards on the object is equal to the weight of the object acting downwards. The buoyant force is also equal to the weight of the fluid displaced by the submerged part of the object.

Weight of cube = Buoyant force

Weight of cube = Mass of cube \(\times\) acceleration due to gravity (\(g\))

Buoyant force = Weight of displaced water = Mass of displaced water \(\times g\)

Mass of displaced water = Volume of displaced water \(\times\) Density of water

So, Mass of cube \(\times g\) = (Volume of displaced water \(\times\) Density of water) \(\times g\)

Since \(g\) is on both sides, we can cancel it out:

Mass of cube = Volume of displaced water \(\times\) Density of water

4. Calculate the volume of displaced water (which is the submerged volume):

Let \(V_{\text{submerged}}\) be the volume of the cube submerged in water. According to the principle, the mass of the cube is equal to the mass of the water displaced by the submerged volume.

\(m_{\text{cube}} = V_{\text{submerged}} \times \rho_{\text{water}}\)

We want to find \(V_{\text{submerged}}\):

\(V_{\text{submerged}} = \frac{m_{\text{cube}}}{\rho_{\text{water}}}\)

Using our values in grams and cm³:

\(V_{\text{submerged}} = \frac{800 \text{ gm}}{1 \text{ gm/cm}^3}\)

\(V_{\text{submerged}} = 800 \text{ cm}^3\)

This is the volume of the cube that is underwater.

5. Calculate the volume of the cube outside the water:

The volume of the cube outside the water is the total volume of the cube minus the submerged volume.

Volume outside = \(V_{\text{total}} - V_{\text{submerged}}\)

Volume outside = \(1000 \text{ cm}^3 - 800 \text{ cm}^3\)

Volume outside = \(200 \text{ cm}^3\)

Therefore, the volume of the cube that is outside the water is 200 cm³.

Revision Table: Floating Cube Calculation

Additional Information: Buoyancy and Floating

Archimedes' Principle: This principle states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

• For a fully submerged object, the displaced volume is the object's total volume.
• For a floating object, the displaced volume is equal to the volume of the object that is submerged in the fluid.

Condition for Floating: An object floats when its average density is less than the density of the fluid it is placed in. In this case, the average density of the cube is:

\(\rho_{\text{cube}} = \frac{\text{Mass}}{\text{Volume}} = \frac{800 \text{ gm}}{1000 \text{ cm}^3} = 0.8 \text{ gm/cm}^3\)

Since \(0.8 \text{ gm/cm}^3 < 1 \text{ gm/cm}^3\) (density of water), the cube floats.

The ratio of submerged volume to total volume is equal to the ratio of the object's density to the fluid's density:

\(\frac{V_{\text{submerged}}}{V_{\text{total}}} = \frac{\rho_{\text{cube}}}{\rho_{\text{water}}}\)

\(\frac{V_{\text{submerged}}}{1000 \text{ cm}^3} = \frac{0.8 \text{ gm/cm}^3}{1 \text{ gm/cm}^3}\)

\(V_{\text{submerged}} = 0.8 \times 1000 \text{ cm}^3 = 800 \text{ cm}^3\)

This confirms our previous calculation for the submerged volume. The volume outside is then \(1000 \text{ cm}^3 - 800 \text{ cm}^3 = 200 \text{ cm}^3\).